### Hyperbola

• Find equation of hyperbola- regular and rectangular, eccentricity, directix, focus, latus rectum, asymptote, relation between a and b.

Equation of a hyperbola when center is at (h,k) $\dfrac{(x-h)^{2}}{a^{2}}- \dfrac{(y-k)^{2}}{b^{2}} = 1$ Equation of a hyperbola when center is at (0,0) $\dfrac{x^{2}}{a^{2}}- \dfrac{y^{2}}{b^{2}} = 1$

• Focus of hyperbola: The two points on the transverse axis. These points are what controls the entire shape of the hyperbola since the hyperbola’s graph is made up of all points, P, such that the difference of distances between P and the two foci are equal. To determine the foci you can use the formula: $a^2 + b^2 = c^2$

• Transverse axis: This is the axis on which the two foci are.

• Asymptotes: The two lines that the hyperbolas come closer and closer to touching. The equation of the asymptotes is always: $y = \pm \frac{a}{b} x$

### Solved Example:

#### 1-6-01

Find the equation of the hyperbola having vertices: ($\pm$5,0), foci: ($\pm$8,0)

Solution:
For hyperbola, $\mathrm{vertices} = (\pm a,0) = (\pm5,0)$,
$\mathrm{foci} = (\pm c,0), c^2 = a^2 + b^2$,
$\mathrm{\ or,\ }b^2 = 8^2 - 5^2 = 64 - 25 = 39$
Equation of a hyperbola, $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 , \dfrac{x^2}{25} - \dfrac{y^2}{39} =1$

### Solved Example:

#### 1-6-02

What is the equation of the asymptote of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{4} =1$?

Solution:
$a = 3, b = 2$. Equation of the asymptote is: $y = \pm \dfrac{b}{a}x, y = \pm \dfrac{2}{3}x, 3y = 2x, 2x - 3y = 0$

### Solved Example:

#### 1-6-03

Find the equation of the hyperbola whose asymptotes are y = $\pm$ 2x and which passes through (5/2, 3).

Solution:
The equation of the asymptote is, $y = \dfrac{b}{a}x$.
Comparing with the given equation,$\dfrac{b}{a} = 2, b=2a$
So the equation of the hyperbola will become, $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{(2a)^{2}} = 1$
Substituting the given co-ordinates, $\dfrac{(5/2)^{2}}{a^{2}}-\dfrac{3^{2}}{(2a)^{2}} = 1, a = 2, b= 4$ $\dfrac{x^{2}}{2^{2}}-\dfrac{y^{2}}{4^{2}} = 1, 4x^2 - y^2 = 16$

### Solved Example:

#### 1-6-04

4$x^2$ - $y^2$ = 16 is the equation of a/an?

Solution:
$4x^{2}-y^{2}=16, \dfrac {x^{2}}{4}-\dfrac {y^{2}}{16}=1$ Because of the negative sign of $y^2$ term this is an equation of a hyperbola in the format, $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$

### Solved Example:

#### 1-6-05

Find the eccentricity of the curve: $9x^2 - 4y^2 - 36x + 8y = 4$

Solution:
\begin{align*} 9x^{2}-36x-4y^{2}+8y &=4\\ 9\left( x^{2}-4x+4\right) -36-4\left( y^{2}-2y+1\right) +4 &=4\\ \dfrac {9}{36}\left( x-2\right) ^{2}-\dfrac {4}{36}\left( y-1\right) ^{2} &=\dfrac {36}{36}\\ \dfrac {\left( x-2\right) ^{2}}{4}-\dfrac {\left( y-1\right) ^{2}}{9} &=1 \end{align*} a = 2, b =3 $b^{2} =a^{2}\left( e^{2}-1\right), 9 =4\left( e^{2}-1\right), 2.25 =e^{2}-1, e =1.80$

How far from the x-axis is the focus F of the hyperbola $x^2 - 2y^2 + 4x + 4y + 4 = 0?$
\begin{align*} x^{2}-2y^{2}+4x+4y+4 &=0\\ x^{2}+4x+4-2y^{2}+4y&=0\\ \left( x+2\right) ^{2}-2\left( y^{2}-2y\right) &=0\\ \left( x+2\right) ^{2}-2\left( y^{2}-2y+1\right) &=-2\\ \left( x+2\right) ^{2}-2\left( y-1\right) ^{2}&=-2\\ 2\left( y-1\right) ^{2}-\left( x+2\right) ^{2}&=2\\ \left( y-1\right) ^{2}-\dfrac {\left( x+2\right) ^{2}}{2}&=1\\ \end{align*} Comparing with, $\dfrac {Y^{2}}{a^{2}}-\dfrac {X^{2}}{b^{2}}=1, a=1, \ b=\sqrt {2}$ $X = x+2\ ,\ Y = y-1$ $e =\dfrac {\sqrt {a^{2}+b^{2}}}{a} =\dfrac {\sqrt {1+2}}{1} =\sqrt {3},$ $F\equiv \left( 0,\pm ae\right) =\left( 0,\pm \sqrt {3}\right)$ $X = x+2 \mathrm{\ Focus\ Shifted\ by}\ -2 \mathrm{\ in\ } x \mathrm{\ axis},$ $Y = y-1 \mathrm{\ Focus\ Shifted\ by}\ 1 \mathrm{\ in\ } y \mathrm{\ axis}$ $\mathrm{Focus\ Shifted\ by}=\left( -2,1\right)$ $F_{1} \equiv \left( -2,1+\sqrt {3}\right) =\left( -2,2.73\right),$ $F_{2}\equiv \left( -2,1-\sqrt {3}\right) =\left( -2,-0.73\right)$